Hi all,
I have a square that lies in a coordinate plane from (-100,100) to (100,-100). I would like to know how to find the corners of an inset square when the outer square is rotated some angle about the center.
Thank you.
Hi all,
I have a square that lies in a coordinate plane from (-100,100) to (100,-100). I would like to know how to find the corners of an inset square when the outer square is rotated some angle about the center.
Thank you.
Hello Aondiar
Welcome to Math Help Forum!I'm not sure that I fully understand the details of your description of the problem, but if you take a look at the attached diagram, you can see whether it's what you want.
The square $\displaystyle ABCD $ has been rotated through an angle $\displaystyle \theta$ anticlockwise, where $\displaystyle 0^o<\theta<90^o$. For instance, $\displaystyle \angle AOA' = \theta$, giving $\displaystyle \angle POA' = \theta - 45^o$. The square has then been reduced in size so that its vertices lie on the perimeter of the original square, to produce $\displaystyle A'B'C'D'$.
Then, as shown in the diagram, $\displaystyle A'$ has coordinates $\displaystyle \big(100, 100\tan(\theta - 45^o)\big)$
The coordinates of the remaining vertices are then fairly obvious.
Let me know if the diagram is not what you meant.
Grandad